Correct Answer - Option 2 : 3

__Concept:__

Let us consider sequence a1, a2, a3 …. an is an A.P.

- Common difference “d”= a2 – a1 = a3 – a2 = …. = an – an – 1

- nth term of the A.P. is given by an = a + (n – 1) d
- Sum of the first n terms = Sn =\(\rm \frac n 2\) [2a + (n − 1) × d]= \(\rm \frac n 2\)(a + l)

Where, a = First term, d = Common difference, n = number of terms, an = nth term and l = Last term

__Calculation:__

Let the first term of AP be 'a' and the common difference be 'd'

Given: Fourth term of an A.P. is zero

⇒ a_{4} = 0

⇒ a + (4 - 1) × d = 0

⇒ a + 3d = 0

∴ a = -3d .... (1)

**To Find:** \(\rm \dfrac{t_{25}}{t_{11}}\)

\(\rm \Rightarrow \dfrac{t_{25}}{t_{11}} = \dfrac {a+24d}{a+10d}\\=\dfrac {-3d+24d}{-3d+10d}\\=\dfrac {21d} {7d}=3\)